Question

Equations with the Dependent Variable Missing. For a scond order differential equation of the form \(y^{\prime \prime}=f\left(t, y^{\prime}\right),\) the substitution \(v=y^{\prime}, v^{\prime}=y^{\prime \prime}\) leads to a first order equation of the form \(v^{\prime}=f(t, v)\). If this equation can be solved for \(v\), then \(y\) can be obtained by integrating \(d y / d t=v .\) Note that one arbitrary constant is obtained in solving the first order equation for \(v,\) and a second is introduced in the integration for \(y\). In each of Problems 28 through 33 use this substitution to solve the given equation. $$ t y_{0}^{\prime \prime}+y^{\prime}=1, \quad t>0 $$

Short answer

The final solution is given by: \(y = t - C_2\frac{t^2}{2} + C_3\), where \(C_2\) and \(C_3\) are arbitrary constants.

Step-by-step solution

Making Substitutions:

Replace \(y^{\prime}\) with \(v\) and \(y^{\prime\prime}\) with \(v^{\prime}\) in the given equation: $$ tv^{\prime}+v = 1, \quad t>0. $$

Solve for \(v^{\prime}\):

Rearrange the equation for \(v^{\prime}\): $$ v^{\prime} = \frac{1-v}{t}. $$

Solve the first order differential equation for \(v\):

Use the separation of variables technique to solve for \(v\): $$ \frac{dv}{1-v} = \frac{dt}{t}. $$ Integrate both sides of the equation: $$ \int \frac{dv}{1-v} = \int \frac{dt}{t}. $$ $$ -\ln|1-v| = \ln|t|+C_1. $$ Now, solve for \(v\): $$ 1-v = \pm te^{-C_1}. $$ Let \(C_2 = \pm e^{-C_1}\), then $$ v = 1 - C_2t. $$

Find \(y\):

Now find \(y\) by integrating the equation \(dy/dt = v\): $$ \frac{dy}{dt}=1-C_2t. $$ Integrate both sides of the equation: $$ \int dy = \int (1-C_2t) dt. $$ $$ y= t-C_2\frac{t^2}{2}+C_3. $$

Write the final solution:

The final solution of the given differential equation is: $$ y= t-C_2\frac{t^2}{2}+C_3, $$ where \(C_2\) and \(C_3\) are arbitrary constants.

Key concepts

First-Order Differential Equations

First-order differential equations involve derivatives of unknown functions and do not go beyond the first derivative. They are simpler than second-order differential equations because they only require finding one integration constant rather than two. In the problem above, the original second-order differential equation is transformed into a first-order differential equation using substitutions. This new equation only involves the first derivative (denoted as \(v\) or \(v'\)), simplifying the process of finding a solution. Solving first-order differential equations often involves techniques like separation of variables, making them approachable even for students who are new to differential equations.

Separation of Variables

Separation of variables is a common method used to solve first-order differential equations. It involves rearranging the equation so that each variable and its differential are on opposite sides of the equation. This makes them integrable separately. In the exercise, after the substitution, we transformed the equation into a form that allowed separation: \(\frac{dv}{1-v} = \frac{dt}{t}\). By integrating each side, we found a solution for \(v\). Here are the general steps for separation of variables:

• Rearrange the equation to isolate the differentials.
• Integrate both sides separately.
• Solve for the unknown function or its derivative.

Apply these steps carefully to separate, integrate, and solve.

Integration Constants

Integration constants arise in the process of solving differential equations. Whenever you integrate, you include a constant of integration, reflecting the general solution that represents a family of curves. In our solved example, two constants of integration, \(C_1\) and \(C_3\), emerged: one from solving the first-order equation for \(v\), and the other from integrating to find \(y\). The presence of integration constants is crucial because they account for all possible initial conditions or specific solutions that a differential equation model might have.Remember:

• Integration constants are added at the end of each integration step.
• General solutions require these constants to accommodate various boundary conditions or initial values.

These constants provide flexibility and adaptability in matching specific scenarios.

Substitution Method

The substitution method is an essential tool in solving differential equations, particularly when the equation is transformed to a simpler form as a result. In the provided exercise, we used the substitution \(v = y'\) and \(v' = y''\) to transform a second-order differential equation into a first-order one. This approach simplifies the equation, making it easier to handle with techniques such as separation of variables. Substitution relies on:

• Choosing the appropriate variables to substitute, simplifying the system.
• Transforming the differential equation into a form that is easier to solve.
• Eventually replacing the substituted variables with original terms to find the final solution.

It's vital to correctly perform the substitution and reverse it after solving the differential equation to answer the original problem.